To solve this problem, we need to determine the average velocity for each leg of Adam's bus trip. The average velocity is found by dividing the distance traveled by the time taken for each leg of the trip. The formula for average velocity is:
[tex]\[ \text{Average Velocity} = \frac{\text{Distance}}{\text{Time}} \][/tex]
Let's calculate the average velocity for each leg (A, B, C, D, E):
Leg A:
- Distance = 18 km
- Time = 9 min
[tex]\[ \text{Average Velocity}_{\text{A}} = \frac{18 \text{ km}}{9 \text{ min}} = 2 \text{ km/min} \][/tex]
Leg B:
- Distance = 25 km
- Time = 15 min
[tex]\[ \text{Average Velocity}_{\text{B}} = \frac{25 \text{ km}}{15 \text{ min}} \approx 1.67 \text{ km/min} \][/tex]
Leg C:
- Distance = 24 km
- Time = 8 min
[tex]\[ \text{Average Velocity}_{\text{C}} = \frac{24 \text{ km}}{8 \text{ min}} = 3 \text{ km/min} \][/tex]
Leg D:
- Distance = 48 km
- Time = 12 min
[tex]\[ \text{Average Velocity}_{\text{D}} = \frac{48 \text{ km}}{12 \text{ min}} = 4 \text{ km/min} \][/tex]
Leg E:
- Distance = 15 km
- Time = 7 min
[tex]\[ \text{Average Velocity}_{\text{E}} = \frac{15 \text{ km}}{7 \text{ min}} \approx 2.14 \text{ km/min} \][/tex]
Next, we will arrange the legs from highest average velocity to lowest average velocity:
1. Leg D: [tex]\(4 \text{ km/min}\)[/tex]
2. Leg C: [tex]\(3 \text{ km/min}\)[/tex]
3. Leg E: [tex]\(\approx 2.14 \text{ km/min}\)[/tex]
4. Leg A: [tex]\(2 \text{ km/min}\)[/tex]
5. Leg B: [tex]\(\approx 1.67 \text{ km/min}\)[/tex]
Therefore, the legs of the trip arranged from highest velocity to lowest are:
[tex]\[ \text{D}, \text{C}, \text{E}, \text{A}, \text{B} \][/tex]
So, the average velocities and the ordered legs of the trip are:
Average Velocities:
- A: [tex]\(2 \text{ km/min}\)[/tex]
- B: [tex]\(\approx 1.67 \text{ km/min}\)[/tex]
- C: [tex]\(3 \text{ km/min}\)[/tex]
- D: [tex]\(4 \text{ km/min}\)[/tex]
- E: [tex]\(\approx 2.14 \text{ km/min}\)[/tex]
Order from highest to lowest velocity:
[tex]\[ \text{D}, \text{C}, \text{E}, \text{A}, \text{B} \][/tex]
